Discrete Math Help: Proving Injectivity of f & g

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SUMMARY

The discussion centers on the injectivity of composite functions in discrete mathematics, specifically addressing the functions f: B => C and g: A => B. It is established that if the composition f(g(x)) is injective, then f must be injective, while the converse is not necessarily true. The proof strategy involves assuming the non-injectivity of either function and deriving a contradiction, thereby confirming the injectivity of f when f(g(x)) is injective.

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  • Understanding of injective functions in mathematics
  • Familiarity with function composition
  • Basic knowledge of proof techniques, particularly proof by contradiction
  • Concepts of discrete mathematics
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  • Study the properties of injective, surjective, and bijective functions
  • Learn about proof techniques in discrete mathematics, focusing on proof by contradiction
  • Explore function composition and its implications in mathematical analysis
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Homework Statement



f: B => C and g: A => B

1. If f of g is injective, then f is injective.

2. If f of g is injective, then g is injective.

Homework Equations



The Attempt at a Solution

I know that 1 is true and 2 is false because I found those as properties, but I am not exactly sure why, and I do not know how to show a proof.

Any help please?
 
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If f of g is injective then every x,y in A such that f(g(x)) = f(g(y)) implies that x=y.

I suggest from here to suppose that f or g is not injective and come up with a contradiction. Ie: suppose f is injective and g is not injective, suppose f is not injective and g is injective, suppose f and g are not injective. Work it out from here.
 

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